priority load shedding circuit...

Discussion in 'The Projects Forum' started by lostinspace73, Jan 21, 2011.

  1. lostinspace73

    Thread Starter New Member

    Jan 21, 2011
    Hey. I'm in the project development stage of a priority load shedding circuit. I have a 120VAC source (3600VA inverter) providing a nominal current of 30A to prioritized loads (whose total draw exceeds the rating of the inverter). I understand that a current transformer is necessary to meter the line current but cannot seem to get my head around the circuitry involved. The CT will be interfaced to a PIC18F4550 which will be responsible for shedding loads according to coded priority via relays (yet to be determined). I plan on powering the PIC with a LP2950 voltage regulator.

    That, unfortunately, is where my very general understanding of the circuit ends. I'm not looking for a schematic. What I feel I really need to get rolling with this is some professional advice on how to approach the project...a block diagram would be ideal.

    120VAC LOADS
    700W Microwave
    1100W Coffeemaker
    700W (peak) Blender
    1250W (Max.) Electric Frying Pan
    ?W-??W Backup/Primary Lighting

    AAC Fanatic!

    Jul 1, 2008
    Another school project for saving the planet perchance? Some years ago I installed an 'On Demand' hot water heater in my home. Not because of any 'Go Green' issue, but just because it freed up an entire utility closet with a wall unit the size of a small town telephone book. Of course, the down side of these water heaters is that they have absolutely no reserve (hot water tank) hot water, with exception of what's in the pipes. Come to think of it, there's nothing I'd like more than a Load shedding circuit that shuts my water heater off while I'm showering because my wife decided to make a pot of coffee. :rolleyes:
  3. kubeek

    AAC Fanatic!

    Sep 20, 2005
  4. beenthere

    Retired Moderator

    Apr 20, 2004
  5. eblc1388

    AAC Fanatic!

    Nov 28, 2008
    Be aware if you want to use the circuit suggested in the link of post #3 above.

    Probably the designer hasn't actually read the data sheet of the MOC3021.

    It has been stated clearly in the MOC3021 data sheet that the input IR emitter current should be somewhere between 15mA to the max. 60mA IR diode rating for correct operation. A good value to use is 20mA.

    With 3.3V drive and an extra 1N4001, plus the nominal 1.15~1.5V voltage drop across the IR emitter, what would be the current in the 470Ω resistor?

    Let's do the maths, assuming 1.3V for the IR drop.

    Current = (3.3 - 0.65 - 1.3) / 470 = 2.9mA

    Even with the drive voltage at +5V, the current is still only 6.5mA, far lower than that needed for reliable operation.

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  6. lostinspace73

    Thread Starter New Member

    Jan 21, 2011
    Thanks everyone. You've helped me focus my research efforts.
  7. Wendy


    Mar 24, 2008
    Most SSRs (Solid State Relays), which are optocoupled TRIACs, already have the resistor. Simple DC in, and the AC side conducts. They are very available, and extremely easy to use, the specs are written right on the case. Think 3V-9V DC inputs.